1. ## Potential difference

If we're given V=2x^2+2y^2 and we're told to find the field at a point x=3, y=3, would this method be correct?

Ex=d(2x^2)/dx=4x=12

Ey=d(-2y^2)/dy=-4y=-12

Therefore, E={12i - 12j}

Not entirely sure, this is the first of its kind.

2. Originally Posted by quantoembryo
If we're given V=2x^2+2y^2 and we're told to find the field at a point x=3, y=3, would this method be correct?

Ex=d(2x^2)/dx=4x=12

Ey=d(-2y^2)/dy=-4y=-12

Therefore, E={12i - 12j}

Not entirely sure, this is the first of its kind.
Mostly correct. $E = - \nabla V$ so you are off by a minus sign.

-Dan

3. So then it would be E={-12i+12j}? That actually makes a lot more sense.

4. Do you have a typo in your potential?

You wrote in your post that: $V=2x^2+2y^2\,.$

You keep responding as if $V=2x^2-2y^2\,.$

5. Good catch. I missed that.

-Dan

6. Yes, that was a typo. Sorry! Thanks for the help! One quick question pertaining to that relationship you posted. What is that greek symbol and what does it mean? Thanks

7. If you are referring to the $\nabla$ in the topsquark's expression $E = - \nabla V,$ then it's not a Greek letter, actually. It's called the "del" operator, or also the "nabla". In this context, it indicates that you are to take the gradient of the scalar function $V.$

8. Alright, I have yet to encounter that sign so it's probably best if I say E=-dv/dx correct?

9. Alright, I have yet to encounter that sign so it's probably best if I say E=-dv/dx correct?
Not true for your problem, actually. If you were working in only one dimension, it would be true. For your case, you should have the vector equation

$\mathbf{E}=-\left\langle\dfrac{\partial V}{\partial x},\dfrac{\partial V}{\partial y},\dfrac{\partial V}{\partial z}\right\rangle.$

Or, for two dimensions, you'd have

$\mathbf{E}=-\left\langle\dfrac{\partial V}{\partial x},\dfrac{\partial V}{\partial y}\right\rangle.$

10. Alright, let me rephrase. Ex=-dV/dx, and Ey=-dV/dy

11. As long as your "d's" mean "partial derivative" and not "ordinary derivative", you're ok there.